55.2 potassium carbonate solution by mass fraction of 25% was mixed with 55.5 g

55.2 potassium carbonate solution by mass fraction of 25% was mixed with 55.5 g of calcium chloride solution with a concentration of 20%. determine the concentration of potassium chloride in the resulting solution.

Given:
m solution (K2CO3) = 55.2 g
ω (K2CO3) = 25%
m solution (CaCl2) = 55.5 g
ω (CaCl2) = 20%

To find:
ω (KCl) -?

Decision:
1) K2CO3 + CaCl2 => 2KCl + CaCO3 ↓;
2) m (K2CO3) = ω (K2CO3) * m solution (K2CO3) / 100% = 25% * 55.2 / 100% = 13.8 g;
3) n (K2CO3) = m (K2CO3) / M (K2CO3) = 13.8 / 138 = 0.1 mol;
4) m (CaCl2) = ω (CaCl2) * m solution (CaCl2) / 100% = 20% * 55.5 / 100% = 11.1 g;
5) n (CaCl2) = m (CaCl2) / M (CaCl2) = 11.1 / 111 = 0.1 mol;
6) n (KCl) = n (CaCl2) * 2 = 0.1 * 2 = 0.2 mol;
7) m (KCl) = n (KCl) * M (KCl) = 0.2 * 74.5 = 14.9 g;
8) n (CaCO3) = n (CaCl2) = 0.1 mol;
9) m (CaCO3) = n (CaCO3) * M (CaCO3) = 0.1 * 100 = 10 g;
10) m solution (KCl) = m solution (K2CO3) + m solution (CaCl2) – m (CaCO3) = 55.2 + 55.5 – 10 = 100.7 g;
11) ω (KCl) = m (KCl) * 100% / m solution (KCl) = 14.9 * 100% / 100.7 = 14.8%.

Answer: Mass fraction of KCl is 14.8%.